Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{3r - 6}{-5r + 45} \times \dfrac{r^2 - 10r + 9}{7r - 7} $
Solution: First factor the quadratic. $x = \dfrac{3r - 6}{-5r + 45} \times \dfrac{(r - 9)(r - 1)}{7r - 7} $ Then factor out any other terms. $x = \dfrac{3(r - 2)}{-5(r - 9)} \times \dfrac{(r - 9)(r - 1)}{7(r - 1)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ 3(r - 2) \times (r - 9)(r - 1) } { -5(r - 9) \times 7(r - 1) } $ $x = \dfrac{ 3(r - 2)(r - 9)(r - 1)}{ -35(r - 9)(r - 1)} $ Notice that $(r - 1)$ and $(r - 9)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 3(r - 2)\cancel{(r - 9)}(r - 1)}{ -35\cancel{(r - 9)}(r - 1)} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $x = \dfrac{ 3(r - 2)\cancel{(r - 9)}\cancel{(r - 1)}}{ -35\cancel{(r - 9)}\cancel{(r - 1)}} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $x = \dfrac{3(r - 2)}{-35} $ $x = \dfrac{-3(r - 2)}{35} ; \space r \neq 9 ; \space r \neq 1 $